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CHAPTER 6 Some discrete distributions 6.1. Examples: Bernoulli, binomial, Poisson, geometric distributions Bernoulli distribution A random variable be a X such that Bernoulli random variable P(X = 1) = p with parameter Var X = p − p2 = p(1 − p). We denote such a random variable by P(X = 0) = 1 − p is said p. Note EX = p and EX 2 = p, and to so X ∼ Bern (p). Binomial distribution A random variable X has a k) = nk pk (1 − p)n−k . binomial distribution We denote such a random variable by The number of successes in n with parameters n and p cumbersome calculations one can derive X = Y1 + · · · + Yn , EX = EY1 + · · · + EYn = np. P(X = X ∼ Binom (n, p). Bernoulli trials is a binomial random variable. is binomial, then if EX = np. An easier way Yi are independent where the After some is to realize that if X Bernoulli variables, so We have not dened yet what it means for random variables to be independent, but here we mean that the events such as (Yi = 1) are independent. Proposition 6.1 X := Y1 + · · · + Yn , where {Yi }ni=1 are independent Bernoulli random variables parameter p, then EX = np, Var X = np(1 − p). Suppose with Proof. First we use the denition of expectation to see that n n X X n i n i n−i EX = i p (1 − p) = i p (1 − p)n−i . i i i=0 i=1 Then 81 82 6. SOME DISCRETE DISTRIBUTIONS EX = n X i i=1 = np n! pi (1 − p)n−i i!(n − i)! n X i=1 (n − 1)! pi−1 (1 − p)(n−1)−(i−1) (i − 1)!((n − 1) − (i − 1))! n−1 X (n − 1)! pi (1 − p)(n−1)−i i!((n − 1) − i)! i=0 n−1 X n − 1 = np pi (1 − p)(n−1)−i = np, i i=0 = np where we used the Binomial Theorem (Theorem 1.1). To get the variance of X, we rst observe that 2 EX = n X i=1 EYi2 + X EYi Yj . i6=j Now EYi Yj = 1 · P(Yi Yj = 1) + 0 · P(Yi Yj = 0) = P(Yi = 1, Yj = 1) = P(Yi = 1)P(Yj = 1) = p2 n 2 2 using independence of random variables {Yi }i=1 . Expanding (Y1 + · · · + Yn ) yields n terms, 2 2 of which n are of the form Yk . So we have n − n terms of the form Yi Yj with i 6= j . Hence Var X = EX 2 − (EX)2 = np + (n2 − n)p2 − (np)2 = np(1 − p). Later we will see that the variance of the sum of independent random variables is the sum of the variances, so we could quickly get Var X = np(1 − p). Alternatively, one can compute E(X 2 ) − EX = E(X(X − 1)) using binomial coecients and derive the variance of X from that. Poisson distribution A random variable X has the Poisson distribution P(X = i) = e−λ We denote such a random variable by X ∼ Pois(λ). ∞ X i=0 so the probabilities add up to one. with parameter λi . i! Note that λi /i! = eλ , λ if 6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONS 83 Proposition 6.2 Suppose X is a Poisson random variable with parameter λ, then EX = λ, Var X = λ. Proof. We start with the expectation EX = ∞ X ie i −λ λ i=0 i! −λ =e ∞ X λi−1 λ = λ. (i − 1)! i=1 Similarly one can show that 2 E(X ) − EX = EX(X − 1) = ∞ X i(i − 1)e−λ i=0 = λ2 e−λ ∞ X i=2 λi i! i−2 λ (i − 2)! 2 =λ , so EX 2 = E(X 2 − X) + EX = λ2 + λ, Example 6.1. and hence Var X = λ. Suppose on average there are 5 homicides per month in a given city. What is the probability there will be at most 1 in a certain month? Solution : If X EX = 5. Since the expectation P(X = 0) + P(X = 1) = e−5 + 5e−5 . is the number of homicides, we are given that for a Poisson is Example 6.2. λ, then λ = 5. Therefore Suppose on average there is one large earthquake per year in California. What's the probability that next year there will be exactly 2 large earthquakes? Solution : λ = EX = 1, so P(X = 2) = e−1 ( 21 ). We have the following proposition connecting binomial and Poisson distributions. Proposition 6.3 (Binomial approximation of Poisson distribution) Xn is a binomial random variable with parameters n and pn P(Xn = i) → P(Y = i), where Y is Poisson with parameter λ. If and npn → λ, then 84 6. SOME DISCRETE DISTRIBUTIONS 6.1 (Approximation of Poisson by binomials) Note that by setting pn := λ/n for n>λ we can approximate the Poisson distribution with parameter tions with parameters n and λ by binomial distribu- pn . This proposition shows that the Poisson distribution models binomials when the probability of a success is small. The number of misprints on a page, the number of automobile accidents, the number of people entering a store, etc. can all be modeled by a Poisson distribution. Proof. For simplicity, let us suppose that can use λn = npn −−−→ λ. n→∞ λ = npn for n > λ. In the general case we We write P(Xn = i) = n! pi (1 − pn )n−i i!(n − i)! n i n−i λ n(n − 1) · · · (n − i + 1) λ 1− = i! n n n i n(n − 1) · · · (n − i + 1) λ (1 − λ/n) = . ni i! (1 − λ/n)i Observe that the following three limits exist n(n − 1) · · · (n − i + 1) −−−→ 1, n→∞ ni i (1 − λ/n) −−−→ 1, n→∞ (1 − λ/n) −−−→ e−λ , n n→∞ which completes the proof. In Section 2.2.3 we considered k = 1, 2, . . . , n. discrete uniform distributions P(X = k) = n1 die (with n = 6), with This is the distribution of the number showing on a for for example. Geometric distribution A random variable X has the geometric distribution with parameter P(X = i) = (1 − p) i−1 p for p, 0 < p < 1, if i = 1, 2, . . . . Using a geometric series sum formula we see that ∞ X ∞ X P(X = i) = (1 − p)i−1 p = i=1 In Bernoulli trials, if we let i=1 X 1 p = 1. 1 − (1 − p) get a heads, then X X will be a geometric X is the rst time we be the rst time we have a success, then random variable. For example, if we toss a coin over and over and will have a geometric distribution. To see this, to have the rst success 6.1. EXAMPLES: BERNOULLI, BINOMIAL, POISSON, GEOMETRIC DISTRIBUTIONS k th occur on the k−1 (1 − p)k−1 p. trial, we have to have success. The probability of that is failures in the rst k−1 85 trials and then a Proposition 6.4 p, 0 < p < 1, If X is a geometric random variable with parameter then 1 EX = , p 1−p , p2 Var X = FX (k) = P (X 6 k) = 1 − (1 − p)k . Proof. We will use ∞ X 1 = nrn−1 (1 − r)2 n=0 which we can show by dierentiating the formula for geometric series 1/(1 − r) = P∞ Then EX = ∞ X i · P(X = i) = i=1 ∞ X 1 1 2 ·p = . p (1 − (1 − p)) i · (1 − p)i−1 p = i=1 Then the variance 2 X 2 ∞ 1 1 Var X = E (X − EX) = E X − = i− · P(X = i) p p i=1 2 To nd the variance we will use another sum. First ∞ X r nrn , = (1 − r)2 n=0 which we can dierentiate to see that ∞ X 1+r = n2 rn−1 . 3 (1 − r) n=1 Then 2 EX = ∞ X 2 i · P(X = i) = i=1 ∞ X i2 · (1 − p)i−1 p = i=1 2−p (1 + (1 − p)) . 3 ·p = p2 (1 − (1 − p)) Thus 2−p Var X = EX − (EX) = − p2 2 2 2 1−p 1 = . p p2 n=0 rn . 86 6. SOME DISCRETE DISTRIBUTIONS The cumulative distribution function (CDF) can be found by using the geometric series sum formula ∞ X 1 − FX (k) = P (X > k) = P(X = i) = i=k+1 ∞ X i−1 (1 − p) i=k+1 (1 − p)k p = (1 − p)k . p= 1 − (1 − p) Negative binomial distribution A random variable X has negative binomial distribution P(X = n) = with parameters r and p if n−1 r p (1 − p)n−r , n = r, r + 1, . . . . r−1 A negative binomial represents the number of trials until r successes. To get the above th th success in the n trial, we must exactly have r − 1 successes in the formula, to have the r th rst n − 1 trials and then a success in the n trial. Hypergeometric distribution A random variable X has hypergeometric distribution with parameters m, n and N if m N −m i n−i . P(X = i) = N n N balls, of which m are one N −m are another, and we choose n balls at random without replacement, the probability of having i balls of the rst color. This comes up in sampling without replacement: if there are color and the other then X represents Another model where the hypergeometric distribution comes up is the probability of a success changes on each draw, since each draw decreases the population, in other words, when we consider sampling without replacement from a nite population). Then size, m is the number of success states in the population, quantity drawn in each trial, i n N is the population is the number of draws, that is, is the number of observed successes. 6.2. FURTHER EXAMPLES AND APPLICATIONS 87 6.2. Further examples and applications 6.2.1. Bernoulli and binomial random variables. Example 6.3. A company prices its hurricane insurance using the following assumptions: (i) In any calendar year, there can be at most one hurricane. (ii) In any calendar year, the probability of a hurricane is 0.05. (iii) The numbers of hurricanes in dierent calendar years are mutually independent. Using the company's assumptions, nd the probability that there are fewer than 3 hurricanes in a 20-year period. Solution : X the number of hurricanes X ∼ Binom (20, 0.05), therefore denote by we see that in a 20-year period. From the assumptions P (X < 3) = P (X 6 2) 20 20 20 1 19 0 20 = (0.05) (0.95) + (0.05) (0.95) + (0.05)2 (0.95)18 0 1 2 = 0.9245. Example 6.4. Phan has a 0.6 probability of making a free throw. Suppose each free throw is independent of the other. If he attempts 10 free throws, what is the probability that he makes at least 2 of them? Solution : If X ∼ Binom (10, 0.6), then P (X > 2) = 1 − P (X = 0) − P (X = 1) 10 10 0 10 =1− (0.6) (0.4) − (0.6)1 (0.4)9 0 1 = 0.998. 6.2.2. The Poisson distribution. Recall that a Poisson distribution models well events that have a low probability and the number of trials is high. For example, the probability of a misprint is small and the number of words in a page is usually a relatively large number compared to the number of misprints. (1) The number of misprints on a random page of a book. (2) The number of people in community that survive to age 100. (3) The number of telephone numbers that are dialed in an average day. (4) The number of customers entering post oce on an average day. Example 6.5. Levi receives an average of two texts every 3 minutes. If we assume that the number of texts is Poisson distributed, what is the probability that he receives ve or more texts in a © 9-minute period? Copyright 2017 Phanuel Mariano, Patricia Alonso Ruiz, Copyright 2020 Masha Gordina. 88 6. SOME DISCRETE DISTRIBUTIONS Solution : Let X be the number of texts in a 9−minute period. Then λ=3·2=6 and P (X > 5) = 1 − P (X 6 4) =1− 4 X 6n e−6 n=0 n! = 1 − 0.285 = 0.715. Example 6.6. tation λ. Solution : Let X1 , ..., Xk be independent Poisson random variables, each with expec- What is the distribution of the random variable The distribution of Y is Poisson with the expectation use Proposition 6.3 and (6.1) to choose pn = kλ1 /n = λ1 /m = λ/n Y := X1 + ... + Xk ? n = mk λ = kλ. To show this, we Bernoulli random variables with parameter to approximation the Poisson random variables. them all together, the limit as n → ∞ If we sum gives us a Poisson distribution with expectation lim npn = λ. However, we can re-arrange the same n = mk Bernoulli random variables n→∞ in k groups, each group having m Bernoulli random variables. Then the limit gives us the distribution of X1 + ... + Xk . This argument can be made rigorous, but this is beyond the scope of this course. Note that we do not show that the we have convergence in distribution. Example 6.7. Let λ1 , . . . , λk , X1 + ... + Xk ? pectation Solution : X1 , . . . , X k be independent Poisson random variables, each with ex- respectively. Y = λ = λ1 + ... + λk . To show this, we again use Proposition 6.3 and (6.1) with parameter pn = λ/n. If n is large, we can separate these n Bernoulli random variables in k groups, each having ni ≈ λi n/λ Bernoulli random variables. The result follows if lim ni /n = λi for each i = 1, ..., k . The distribution of Y What is the distribution of the random variable is Poisson with expectation n→∞ This entire set-up, which is quite common, involves what is called distributed Bernoulli random variables Example 6.8. independent identically (i.i.d. Bernoulli r.v.). Can we use binomial approximation to nd the mean and the variance of a Poisson random variable? Solution : Yes, and this is really simple. Recall again from Proposition 6.3 and (6.1) that we can approximate Poisson where pn = λ/n. Y with parameter λ by a binomial random variable Each such a binomial random variable is a sum on random variables with parameter pn . n→∞ n independent Bernoulli Therefore EY = lim npn = lim n n→∞ λ = λ, n λ Var(Y ) = lim npn (1 − pn ) = lim n n→∞ n→∞ n Binom (n, pn ), λ 1− = λ. n 6.2. FURTHER EXAMPLES AND APPLICATIONS 6.2.3. Table of distributions. The following table summarizes the discrete distribu- tions we have seen in this chapter. N0 = N ∪ {0} Here stands for the set of positive integers, and N is the set of nonnegative integers. PMF (k Name Notation Parameters Bernoulli Bern(p) p ∈ [0, 1] 1 k Binomial Binom(n, p) n∈N p ∈ [0, 1] n k Poisson Pois(λ) λ>0 Geometric Geo(p) p ∈ (0, 1) Negative NBin(r, p) r∈N p ∈ (0, 1) binomial Hypergeometric 89 Hyp(N, m, n) N ∈ N0 n, m ∈ N0 ∈ N0 ) pk (1 − p)1−k pk (1 − p)n−k E[X] Var(X) p p(1 − p) np np(1 − p) k e−λ λk! λ ( (1 − p)k−1 p, for k > 1, 1 p 0, else. ( k−1 r p (1 − p)k−r , if k ≥ r, r r−1 p 0, else. −m (mk)(Nn−k ) N (n) nm N λ 1−p p2 r(1−p) p2 nm(N −n) m (1−N ) N (N −1) 90 6. SOME DISCRETE DISTRIBUTIONS 6.3. Exercises Exercise 6.1. A UConn student claims that she can distinguish Dairy Bar ice cream from Friendly's ice cream. As a test, she is given ten samples of ice cream (each sample is either from the Dairy Bar or Friendly's) and asked to identify each one. She is right eight times. What is the probability that she would be right exactly eight times if she guessed randomly for each sample? Exercise 6.2. A Pharmaceutical company conducted a study on a new drug that is sup- posed to treat patients suering from a certain disease. The study concluded that the drug did not help 25% of those who participated in the study. What is the probability that of 6 randomly selected patients, 4 will recover? Exercise 6.3. 20% of all students are left-handed. A class of size 20 meets in a room with 18 right-handed desks and 5 left-handed desks. What is the probability that every student will have a suitable desk? Exercise 6.4. A ball is drawn from an urn containing 4 blue and 5 red balls. After the ball is drawn, it is replaced and another ball is drawn. Suppose this process is done 7 times. (a) What is the probability that exactly 2 red balls were drawn in the 7 draws? (b) What is the probability that at least 3 blue balls were drawn in the 7 draws? Exercise 6.5. The expected number of typos on a page of the new Harry Potter book is 0.2. What is the probability that the next page you read contains (a) 0 typos? (b) 2 or more typos? (c) Explain what assumptions you used. Exercise 6.6. The monthly average number of car crashes in Storrs, CT is 3.5. What is the probability that there will be (a) at least 2 accidents in the next month? (b) at most 1 accident in the next month? (c) Explain what assumptions you used. Exercise 6.7. Suppose that, some time in a distant future, the average number of bur- glaries in New York City in a week is 2.2. Approximate the probability that there will be (a) no burglaries in the next week; (b) at least 2 burglaries in the next week. Exercise 6.8. The number of accidents per working week in a particular shipyard is Poisson distributed with mean 0.5. Find the probability that: (a) In a particular week there will be at least 2 accidents. 6.3. EXERCISES 91 (b) In a particular two week period there will be exactly 5 accidents. (c) In a particular month (i.e. 4 week period) there will be exactly 2 accidents. Exercise 6.9. Jennifer is baking cookies. She mixes 400 raisins and 600 chocolate chips into her cookie dough and ends up with 500 cookies. (a) Find the probability that a randomly picked cookie will have three raisins in it. (b) Find the probability that a randomly picked cookie will have at least one chocolate chip in it. (c) Find the probability that a randomly picked cookie will have no more than two bits in it (a bit is either a raisin or a chocolate chip). Exercise 6.10. A roulette wheel has 38 numbers on it: the numbers 0 through 36 and a 00. Suppose that Lauren always bets that the outcome will be a number between 1 and 18 (including 1 and 18). (a) What is the probability that Lauren will lose her rst 6 bets. (b) What is the probability that Lauren will rst win on her sixth bet? Exercise 6.11. In the US, albinism occurs in about one in 17,000 births. Estimate the probabilities no albino person, of at least one, or more than one albino at a football game with 5,000 attendants. Use the Poisson approximation to the binomial to estimate the probability. Exercise 6.12. An egg carton contains 20 eggs, of which 3 have a double yolk. To make a pancake, 5 eggs from the carton are picked at random. What is the probability that at least 2 of them have a double yolk? Exercise 6.13. Around 30,000 couples married this year in CT. Approximate the proba- bility that at least in one of these couples (a) both partners have birthday on January 1st. (b) both partners celebrate birthday in the same month. Exercise 6.14. A telecommunications company has discovered that users are three times as likely to make two-minute calls as to make four-minute calls. The length of a typical call (in minutes) has a Poisson distribution. Find the expected length (in minutes) of a typical call. 92 6. SOME DISCRETE DISTRIBUTIONS 6.4. Selected solutions Solution to Exercise 6.1: This should be modeled using a binomial random variable X, since there is a sequence of trials with the same probability of success in each one. If 1 she guesses randomly for each sample, the probability that she will be right each time is . 2 Therefore, 8 2 1 45 10 1 = 10 . P (X = 8) = 2 2 2 8 Solution to Exercise 6.2: Solution to Exercise 6.3: 6 4 (0.75)4 (0.25)2 For each student to have the kind of desk he or she prefers, there must be no more than 18 right-handed students and no more than 5 left-handed students, so the number of left-handed students must be between 2 and 5 (inclusive). This means that we want the probability that there will be 2, 3, 4, or 5 left-handed students. We use the binomial distribution and get 5 i 20−i X 20 1 4 i=2 i 5 5 . Solution to Exercise 6.4(A): 2 5 7 5 4 2 9 9 Solution to Exercise 6.4(B): P (X > 3) = 1 − P (X 6 2) 0 7 1 6 2 5 5 7 4 5 7 4 5 7 4 − − =1− 9 9 1 9 9 2 9 9 0 Solution to Exercise 6.5(A): e−0.2 Solution to Exercise 6.5(B): 1 − e−0.2 − 0.2e−0.2 = 1 − 1.2e−0.2 . Solution to Exercise 6.5(C): Since each word has a small probability of being a typo, the number of typos should be approximately Poisson distributed. Solution to Exercise 6.6(A): 1 − e−3.5 − 3.5e−3.5 = 1 − 4.5e−3.5 Solution to Exercise 6.6(B): 4.5e−3.5 Solution to Exercise 6.6(C): Since each accident has a small probability it seems reasonable to suppose that the number of car accidents is approximately Poisson distributed. Solution to Exercise 6.7(A): e−2.2 Solution to Exercise 6.7(B): 1 − e−2.2 − 2.2e−2.2 = 1 − 3.2e−2.2 . 6.4. SELECTED SOLUTIONS 93 Solution to Exercise 6.8(A): We have P (X > 2) = 1 − P (X 6 1) = 1 − e−0.5 (0.5)0 (0.5)1 − e−0.5 . 0! 1! Solution to Exercise 6.8(B): In two weeks the average number of accidents will be λ = 0.5 + 0.5 = 1. Then 5 P (X = 5) = e−1 15! . Solution to Exercise 6.8(C): In a 4 week period the average number of accidents will be λ = 4 · (0.5) = 2. Then 2 P (X = 2) = e−2 22! . Solution to Exercise 6.9(A): number of raisins per cookie is 3 −0.8 (0.8) which is e ≈ 0.0383. 3! This calls for a Poisson random variable 0.8, Solution to Exercise 6.9(B): so we take this as our which is 1.2, so we take this 0 1 − P (C = 0) = 1 − e−1.2 (1.2) ≈ 0.6988. 0! Solution to Exercise 6.9(C): The average . We are asking for This calls for a Poisson random variable number of chocolate chips per cookie is P (C > 1), λ R. as our λ. C. P(R = 3), The average We are asking for This calls for a Poisson random variable B. The average 0.8 + 1.2 = 2, so we take this as our λ. We are asking for 0 1 2 P (B = 0) + P (B = 1) + P (B = 2) = e−2 20! + e−2 21! + e−2 22! ≈ .6767. number of bits per cookie is P (B 6 2), which is Solution to Exercise 6.10(A): 1 − 18 38 6 Solution to Exercise 6.10(B): 1 − 18 38 5 18 38 Solution to Exercise 6.11 Let X denote the number of albinos at the game. We have that X ∼ Binom(5000, p) with p = 1/17000 ≈ 0.00029. The binomial distribution gives us 5000 5000 P(X = 0) = 16999 ≈ 0.745 P(X > 1) = 1 − P(X = 0) = 1 − 16999 ≈ 0.255 17000 17000 P(X > 1) = P(X > 1) − P(X = 1) = =1− 16999 5000 17000 5000 − 1 16999 4999 17000 Approximating the distribution of X 1 1 17000 ≈ 0.035633 by a Poisson with parameter λ= 5000 17000 = 5 gives 17 5 P(Y = 0) = exp − 17 ≈ 0.745 5 P(Y > 1) = 1 − P(Y = 0) = 1 − exp − 17 ≈ 0.255 5 5 5 P(Y > 1) = P(Y > 1) − P(Y = 1) = 1 − exp − 17 − exp − 17 ≈ 0.035638 17 Solution to Exercise 6.12: Let X be the random variable that denotes the number of X ∼ Hyp(20, 3, 5) and we 17 17 3 3 · · P(X > 2) = P(X = 2) + P(X = 3) = 2 203 + 3 202 . eggs with double yolk in the set of chosen 5. Then 5 Solution to Exercise 6.13: We will use Poisson approximation. 5 have that 94 6. SOME DISCRETE DISTRIBUTIONS 1 . If X 3652 denotes the number of married couples where this is the case, we can approximate the −2 distribution of X by a Poisson with parameter λ = 30, 000 · 365 ≈ 0.2251. Hence, −0.2251 P(X > 1) = 1 − P(X = 0) = 1 − e . (a) The probability that both partners have birthday on January 1st is p = (b) In this case, the probability of both partners celebrating birthday in the same month 1/12 and therefore we approximate the distribution by a Poisson λ = 30, 000/12 = 2500. Thus, P(X > 1) = 1 − P(X = 0) = 1 − e−2500 . is Solution to Exercise 6.14: tion, X ∼ Pois(λ) for E[X] = λ. In addition, Let X denote the duration (in minutes) of a call. By assump- λ > 0, so that the expected duration P(X = 2) = 3P(X = 4), which means some parameter we know that λ2 λ4 = 3e−λ . 2! 4! hence E[X] = λ = 2. e−λ From here we deduce that λ2 = 4 with parameter and of a call is Part 2 Continuous random variables CHAPTER 7 Continuous distributions 7.1. Basic theory 7.1.1. Denition, PDF, CDF. We start with the denition a continuous random variable. Denition (Continuous random variables) X is said to have f = fX such that A random variable negative function a continuous distribution if there exists a non- b P(a 6 X 6 b) = f (x)dx a for every a and b. f The function is called the density function for X or the PDF for X. More precisely, such an ∞ f (x)dx = P(−∞ < −∞ Example 7.1. ∞ −∞ f (x)dx = 1 X is said to have an absolutely continuous a distribution. Note that X < ∞) = 1. In particular, P(X = a) = a f (x)dx = 0 for every a. Suppose we are given that and f (x) = c/x3 ∞ f (x)dx = c c −∞ we have ∞ 1 for x>1 and 0 otherwise. Since 1 c dx = , 3 x 2 c = 2. PMF or PDF? Probability mass function (PMF) and (probability) density function (PDF) are two names for the same notion in the case of discrete random variables. We say PDF simply a only for density function for a general random variable, and we use discrete random variables. Denition (Cumulative distribution function (CDF)) The distribution function of X is dened as y F (y) = FX (y) := P(−∞ < X 6 y) = It is also called the cumulative distribution function 97 f (x)dx. −∞ (CDF) of X. PMF or 98 7. CONTINUOUS DISTRIBUTIONS We can dene CDF for any random variable, not just continuous ones, by setting P(X 6 y). F (y) := Recall that we introduced it in Denition 5.3 for discrete random variables. In that case it is not particularly useful, although it does serve to unify discrete and continuous random variables. provided f In the continuous case, the fundamental theorem of calculus tells us, satises some conditions, that f (y) = F 0 (y) . By analogy with the discrete case, we dene the expectation of a continuous random variable. 7.1.2. Expectation, discrete approximation to continuous random variables. Denition (Expectation) For a continuous random variable tation X with the density function by f we dene its expec- ∞ xf (x)dx EX = −∞ if this integral is absolutely convergent. In this case we call X integrable. Recall that this integral is absolutely convergent if ∞ |x|f (x)dx < ∞. −∞ In the example above, ∞ EX = 1 2 x 3 dx = 2 x ∞ x−2 dx = 2. 1 Later in Example 10.1 we will see that a continuous random variable with Cauchy distribution has innite expectation. Proposition 7.1 (Discrete approximation to continuous random variables) Suppose X is a nonnegative continuous random variable with a nite expectation. {Xn }∞ n=1 such that Then there is a sequence of discrete random variables EXn −−−→ EX. n→∞ Proof. First observe that if a continuous random variable density f (x) = 0 x < 0. In particular, F (y) = 0 for y 6 0, X is nonnegative, then its thought the latter is not needed for our proof. Thus for such a random variable ∞ EX = xf (x)dx. 0 n ∈ N, then we dene Xn (ω) to be k/2n if k/2n 6 X(ω) < (k+1)/2n , for k ∈ N∪{0}. −n This means that we are approximating X from below by the largest multiple of 2 that is still below the value of X . Each Xn is discrete, and Xn increase to X for each ω ∈ S . Suppose 7.1. BASIC THEORY Consider the sequence {EXn }∞ n=1 . 99 This sequence is an increasing sequence of positive num- bers, and therefore it has a limit, possibly innite. We want to show that it is nite and it is equal to EX . We have ∞ X k k EXn = P Xn = n 2n 2 k=1 ∞ X k k+1 k = P n 6X< 2n 2 2n k=1 (k+1)/2n ∞ X k = f (x)dx n 2 n k/2 k=1 = ∞ X k=1 If x ∈ [k/2n , (k + 1)/2n ), then x k/2n diers from k f (x)dx. 2n k/2n (k+1)/2n by at most (k+1)/2n xf (x)dx − 06 (k+1)/2n k/2n (k+1)/2n = k/2n k/2n 1/2n , and therefore k f (x)dx 2n (k+1)/2n k 1 f (x)dx x − n f (x)dx 6 n 2 2 k/2n Note that ∞ X k=1 (k+1)/2n ∞ xf (x)dx = k/2n xf (x)dx 0 and (k+1)/2n ∞ n ∞ ∞ X 1 1 1 X (k+1)/2 1 f (x)dx = n f (x)dx = n f (x)dx = n . n 2 k/2n 2 k=1 k/2n 2 0 2 k=1 Therefore 100 7. CONTINUOUS DISTRIBUTIONS ∞ xf (x)dx − 0 6 EX − EXn = 0 = = 6 ∞ X k=1 xf (x)dx − k/2n k=1 ∞ X k=1 ∞ X (k+1)/2n ∞ X k=1 (k+1)/2n ∞ X k=1 (k+1)/2n k/2n (k+1)/2n k/2n (k+1)/2n k/2n 1 2n k/2n (k+1)/2n f (x)dx = k/2n k f (x)dx 2n ! k f (x)dx 2n xf (x)dx − k f (x)dx 2n 1 −−→ 0. 2n n→0 We will not prove the following, but it is an interesting exercise: if discrete random variables that increase up to X , then limm→∞ EXm Xm is any sequence of will have the same value EX . This fact is useful to show linearity, if expectations, then we can take up to Y. Then Xm + Ym Xm X and Y are positive random variables with nite X and Ym discrete X + Y , so we have discrete increasing up to is discrete and increases up to increasing E(X + Y ) = lim E(Xm + Ym ) m→∞ = lim EXm + lim EYm = EX + EY. m→∞ m→∞ Note that we can not easily use the approximations to previous proof to use in this argument, since Xm + Ym X, Y and X+Y we used in the might not be an approximation of the same kind. If X is not necessarily positive, we can show a similar result; we will not do the details. Similarly to the discrete case, we have Proposition 7.2 Suppose X is a continuous random variable with density fX and g is a real-valued function, then ∞ Eg(X) = g(x)f (x)dx −∞ as long as the expectation of the random variable g (X) makes sense. As in the discrete case, this allows us to dene moments, and in particular the Var X := E[X − EX]2 . As an example of these calculations, let us look at the uniform distribution. variance 7.1. BASIC THEORY 101 Uniform distribution We say that a random variable if a6x6b X has a uniform distribution on [a, b] if and 0 otherwise. To calculate the expectation of X ∞ b x xfX (x)dx = EX = −∞ a 1 dx b−a b 1 x dx b−a a 1 b 2 a2 a + b − = . = b−a 2 2 2 = This is what one would expect. To calculate the variance, we rst calculate ∞ b x2 2 2 x fX (x)dx = EX = −∞ a 1 a2 + ab + b2 dx = . b−a 3 We then do some algebra to obtain Var X = EX 2 − (EX)2 = (b − a)2 . 12 fX (x) = 1 b−a 102 7. CONTINUOUS DISTRIBUTIONS 7.2. Further examples and applications Example 7.2. Suppose X has the following p.d.f. ( f (x) = Find the CDF of Solution : X, we have that is, nd FX (x) = 0 if FX (x). x61 2 x3 x>1 x < 1. 0 Use the CDF to nd and will need to compute x FX (x) = P (X 6 x) = 1 x > 1. when P (3 6 X 6 4). 2 1 dy = 1 − 2 3 y x We can use this formula to nd the following probability P (3 6 X 6 4) = P (X 6 4) − P (X < 3) 1 1 7 = FX (4) − FX (3) = 1 − 2 − 1 − 2 = . 4 3 144 Example 7.3. Suppose X has density ( 2x 0 6 x 6 1 fX (x) = . 0 otherwise Find EX . Solution : we have that E [X] = xfX (x)dx = 0 Example 7.4. The density of X 2 x · 2x dx = . 3 is given by ( fX (x) = Find 1 1 2 if 0 otherwise 06x62 . E eX . Solution : using Proposition 7.2 with g(x) = ex 2 EeX = 0 Example 7.5. Suppose X we have 1 1 2 ex · dx = e −1 . 2 2 has density ( 2x 0 6 x 6 1 f (x) = . 0 otherwise © Copyright 2017 Phanuel Mariano, Patricia Alonso Ruiz, Copyright 2020 Masha Gordina. 7.2. FURTHER EXAMPLES AND APPLICATIONS Find 103 Var(X). Solution : in Example 7.3 we found E X 2 E [X] = 2 . Now 3 1 x · 2xdx = 2 = 0 0 Thus 1 Var(X) = − 2 Example 7.6. Suppose X 1 2 1 x3 dx = . 2 2 2 1 = . 3 18 has density ( ax + b 0 6 x 6 1 f (x) = . 0 otherwise and that Solution : E [X 2 ] = 1 . Find the values of 6 We need to use the fact that us a and ∞ −∞ b. f (x)dx = 1 1 1= (ax + b) dx = 0 and the second one give us 1 = 6 x2 (ax + b) dx = a b + . 4 3 Solving these equations gives us a = −2, and E [X 2 ] = a + b, 2 1 0 and b = 2. 1 . The rst one gives 6 104 7. CONTINUOUS DISTRIBUTIONS 7.3. Exercises Exercise 7.1. Let X be a random variable with probability density function ( cx (5 − x) 0 6 x 6 5, f (x) = 0 otherwise. (A) What is the value of c? X? P (2 6 X ≤ 3). (B) What is the cumulative distribution function of (C) Use your answer in part (b) to nd (D) What is (E) What is That is, nd FX (x) = P (X 6 x). E [X]? Var(X)? Exercise 7.2. UConn students have designed the new U-phone. X that the lifetime of a U-Phone is given by the random variable They have determined (measured in hours), with probability density function ( f (x) = 10 x2 x > 10, x ≤ 10. 0 (A) Find the probability that the u-phone will last more than (B) What is the cumulative distribution function of (C) Use part (b) to help you nd Exercise 7.3. Suppose the random variable f (x) = That is, nd X 2 x2 0 has a density function x > 2, x 6 2. An insurance company insures a large number of homes. The insured value, of a randomly selected home is assumed to follow a distribution with density function ( f (x) = 3 x4 x > 1, 0 otherwise. Given that a randomly selected home is insured for at least that it is insured for less than Exercise 7.5. If FX (x) = P (X 6 x). E [X]. Exercise 7.4. X, hours. P (X > 35)? ( Compute X? 20 E [X] = 2. X is given by ( a + bx2 0 6 x 6 1, f (x) = 0 otherwise. The density function of 7 , nd the values of 10 a and b. 1.5, calculate the probability 7.3. EXERCISES Exercise 7.6. Let X be a random variable with density function ( f (x) = Suppose that E [X] = 6 Var(X). Exercise 7.7. 105 1 a−1 0 1 < x < a, otherwise. Find the value of a. Suppose you order a pizza from your favorite pizzeria at 7:00 pm, knowing that the time it takes for your pizza to be ready is uniformly distributed between 7:00 pm and 7:30 pm. (A) What is the probability that you will have to wait longer than 10 minutes for your pizza? (B) If at 7:15pm, the pizza has not yet arrived, what is the probability that you will have to wait at least an additional Exercise 7.8. 10 minutes? The grade of deterioration X of a machine part has a continuous distribution on the interval (0, 10) with probability density function fX (x), where fX (x) is proportional x to on the interval. The reparation costs of this part are modeled by a random variable Y 5 2 that is given by Y = 3X . Compute the expected cost of reparation of the machine part. Exercise 7.9. and 10 : 20, A bus arrives at some (random) time uniformly distributed between and you arrive at a bus stop at 10 : 00 10 : 05. (A) What is the probability that you have to wait at least (B) What is the probability that you have to wait at least 5 minutes until the bus comes? 5 minutes, given that when you arrive today to the station the bus was not there yet (you are lucky today)? Exercise∗ 7.1. For a continuous random variable X with nite rst and second moments prove that E (aX + b) = aEX + b, Var (aX + b) = a2 Var X. for any a, b ∈ R. Exercise∗ 7.2. Let X be a continuous random variable with probability density function x 1 fX (x) = xe− 2 1[0,∞) (x) , 4 where the indicator function is dened as 1[0,∞) (x) = 1, 0 6 x < ∞; 0, Check that fX otherwise. is a valid probability density function, and nd E (X) if it exists. 106 7. CONTINUOUS DISTRIBUTIONS Exercise∗ 7.3. Let X be a continuous random variable with probability density function fX (x) = 4 ln x 1[1,∞) (x) , x3 where the indicator function is dened as 1[1,∞) (x) = 1, 1 6 x < ∞; 0, Check that fX otherwise. is a valid probability density function, and nd E (X) if it exists. 7.4. SELECTED SOLUTIONS 107 7.4. Selected solutions Solution to Exercise 7.1(A): We must have that ∞ f (x)dx = 1, −∞ 5 5 5x2 x3 1= cx(5 − x)dx = c − 2 3 0 0 thus c = 6/125. and so we must have that Solution to Exercise 7.1(B): We have that x f (y)dy FX (x) = P (X 6 x) = −∞ x 6 6 = y (5 − y) dx = 125 0 125 2 3 6 5x x = − . 125 2 3 5y 2 y 3 − 2 3 x 0 Solution to Exercise 7.1(C): We have P (2 6 X 6 3) = P (X 6 3) − P (X < 2) 6 5 · 32 33 5 · 22 23 6 = − − − 125 2 3 125 2 3 = 0.296. Solution to Exercise 7.1(D): we have ∞ E [X] = 5 x· xfX (x)dx = −∞ 0 6 x(5 − x)dx 125 = 2.5. Solution to Exercise 7.1(E): We need to rst compute E X 2 ∞ 5 2 = x2 · x fX (x)dx = −∞ 0 6 x(5 − x)dx 125 = 7.5. Then Var(X) = E X 2 − (E [X])2 = 7.5 − (2.5)2 = 1.25. Solution to Exercise 7.2(A): We have ∞ 20 10 1 dx = . 2 x 2 Solution to Exercise 7.2(B): We have x F (x) = P(X 6 x) = 10 for x > 10, and F (x) = 0 for x < 10. 10 10 dy = 1 − 2 y x 108 7. CONTINUOUS DISTRIBUTIONS Solution to Exercise 7.2(C): We have P (X > 35) = 1 − P (X < 35) = 1 − FX (35) 10 10 =1− 1− = . 35 35 Solution to Exercise 7.3: +∞ Solution to Exercise 7.4: Solution to Exercise 7.5: 37 . 64 we need to use the fact that The rst one gives us 1 0 7 = 10 1 0 −∞ f (x)dx = 1 and E [X] = 7 . 10 b a + bx2 dx = a + 3 1= and the second one gives ∞ a b x a + bx2 dx = + . 2 4 Solving these equations gives 1 a= , 5 Solution to Exercise 7.6: and b= 12 . 5 Note that EX = 1 a x 1 1 dx = a + . a−1 2 2 Also Var(X) = EX 2 − (EX)2 then we need a 2 EX = 1 1 1 1 x2 dx = a2 + a + . a−1 3 3 3 Then 2 1 2 1 1 1 1 Var(X) = a + a+ a+ − 3 3 3 2 2 1 1 1 = a2 − a + . 12 6 12 E [X] = 6 Var(X), we simplify and get 12 a2 − 32 a = 0, which Then, using gives us a = 3. Another way to solve this problem is to note that, for the uniform distribution on [a, b], (a−b)2 (a−1)2 a+b the mean is and the variance is . This gives us an equation 6 = a+1 . Hence 2 12 12 2 2 (a − 1) = a + 1, which implies a = 3. Solution to Exercise 7.7(A): Note that X is uniformly distributed over (0, 30). Then (0, 30). Then 2 P(X > 10) = . 3 Solution to Exercise 7.7(B): Note that X P(X > 25 | X > 15) = is uniformly distributed over P (X > 25) 5/30 = = 1/3. P(X > 15) 15/30 7.4. SELECTED SOLUTIONS Solution to Exercise 7.8: First of all we need to nd the PDF of ( f (x) = Since we have c= 109 cx 5 0 10 X. So far we know that 0 6 x 6 10, otherwise. x c dx = 10c, 5 0 1 . Now, applying Proposition 7.2 we get 10 10 3 3 EY = 0 50 x dx = 150. Solution to Exercise 7.9(A): The probability that you have to wait at least 5 minutes 1 1 until the bus comes is . Note that with probability you have to wait less than 5 minutes, 2 4 1 you already missed the bus. and with probability 4 Solution to Exercise 7.9(B): The conditional probability is 2 . 3